Structure coefficients of a coframe

Question

From the textbook "Introduction to General Relativity, Black Holes & Cosmology" by Yvonne Choquet-Bruhat, p.10:

In a moving coframe in a domain $U$, the differentials of 1-forms $\theta^i$ are given by:

$$d\theta^i \equiv -\cfrac{1}{2} C^i_{jk} \theta^j \wedge \theta^k$$

Show that the structure coefficients of a coframe $\theta^i := a^i_j dx^j$ are given by:

$$ C^i_{hk} \equiv A^j_k\partial_ha^i_j-A^j_h\partial_k a^i_j$$

Where $A$ is the inverse matrix of $a$.

$\underline{\text{An attempt at the solution:}}$

$$ d\theta^i = da^i_j \wedge dx^j \\ = \frac{\partial a^i_j}{\partial x^k} dx^k \wedge dx^j \\ = -\frac{\partial a^i_j}{\partial x^k} dx^j \wedge dx^k \\ = - \frac{\partial a^i_j}{\partial x^k} (A^j_l \theta^l) \wedge (A^k_m \theta^m ) \\ = - [\frac{\partial a^i_j}{\partial x^k}A^j_lA^k_m]\theta^l\wedge\theta^m \\ \implies C^i_{lm} = 2 \frac{\partial a^i_j}{\partial x^k}A^j_lA^k_m $$

Which is clearly incorrect as I only have a single term in my result; where is the error in my working?

Answer 1

Nothing wrong with your calculation. In the book, the author distinguish $\partial_i$ with $\frac{\partial}{\partial x^i}$, as follows : While $\frac{\partial}{\partial x^i}$ is the usual partial derivatives w.r.t. $x^i$ coordinate, the symbol $\partial_i$ is used as the components of exterior derivatives of a function when we express $df$ in coframe $\theta^i$, she called $\partial_i$ as "The Pfaffian Derivatives". That is (in the same page, before equation $I.4.2$) $$ df \equiv \partial_if \, \theta^i. $$ So now you have a choice to find $df$. First as you do $df = \frac{\partial f}{\partial x^i} dx^i$, or you can use $df = \partial_if \, \theta^i$. So now $$ d\theta^i = d(a^i_j dx^j) = da^i_j \wedge dx^j = (\partial_ka^i_j \, \theta^k \wedge dx^j) = \partial_ka^i_j \, \theta^k \wedge (A^j_l \theta^l) = A^j_l \partial_ka^i_j \, \theta^k \wedge \theta^l. $$ As @JohnMa said, the expression $d\theta^i = -\frac{1}{2}C^i_{kl} \, \theta^k \wedge \theta^l $ is in the antisymmetrized form of $d\theta^i$. So we take the antisymmetrization for $d\theta^i = A^j_l \partial_ka^i_j \, \theta^k \wedge \theta^l$ as \begin{align} d\theta^i &= \frac{1}{2}(A^j_l \partial_ka^i_j - A^j_k \partial_la^i_j) \theta^k \wedge \theta^l + \frac{1}{2}(A^j_k \partial_la^i_j-A^j_l \partial_ka^i_j ) \, \theta^l \wedge \theta^k \\ &= -\frac{1}{2}C^i_{kl} \, \theta^k \wedge \theta^l \end{align} So $$ -\frac{1}{2}C^i_{kl} = \frac{1}{2}(A^j_l \partial_ka^i_j - A^j_k \partial_la^i_j) \implies C^i_{kl} = A^j_k \partial_la^i_j - A^j_l \partial_ka^i_j $$ Note that this result $C^i_{hk}=A^j_h \partial_ka^i_j - A^j_k \partial_ha^i_j$ is different (reversed $h$ and $k$) with the one that given on the book, that is $C^i_{hk}=A^j_k \partial_ha^i_j - A^j_h \partial_ka^i_j$. If you're going through earlier computation in the book, about $d^2f$, you'll find that the eq. given in the book $C^i_{hk}=A^j_k \partial_ha^i_j - A^j_h \partial_ka^i_j$ is not match, which means that probably there are typos (indices $h,k$ are reversed) there. It should be $C^i_{hk}=A^j_h \partial_ka^i_j - A^j_k \partial_ha^i_j$.

Answer 2

It is commonly understood that when one writes (in general for a two form)

$$ \alpha = \alpha_{ij} \theta^i \wedge \theta^j \ \ \left( = \sum_{i,j} \alpha_{ij} \theta^i \wedge \theta^j\right), $$

it is assumed that $\alpha_{ij}$ satisfies $\alpha_{ij} = -\alpha_{ji}$. Indeed, if not then the "coefficient" $\alpha_{ij}$ is not well-defined: for example, $\alpha = \theta^1\wedge \theta^2$ can also be written as

$$\alpha = 3 \theta^1 \wedge \theta^2 + 2 \theta^2 \wedge \theta^1.$$

that would give you different $\alpha_{ij}$.

That is, in order to have a meaningful "coefficient", we will always anti-symmetrizes them: we write

\begin{align} \alpha &= b_{ij} \theta^i \wedge \theta^j + b_{ji} \theta^j \wedge \theta^i \\ &= (b_{ij} - b_{ji})\theta^i \wedge \theta^j \\ &= \left( \frac{b_{ij} - b_{ji}}{2}\right)\theta^i\wedge \theta^j + \left( \frac{b_{ji} - b_{ij}}{2}\right)\theta^j \wedge \theta^i. \end{align}

In your situation,

$$ b_{lm} = -\frac{\partial a^i_j}{\partial x^k}A^j_lA^k_m$$

so you should get

$$-\frac 12 C^i_{lm} = \frac{b_{lm} - b_{ml}}{2}=-\frac{1}{2} \left(\frac{\partial a^i_j}{\partial x^k}A^j_lA^k_m - \frac{\partial a^i_j}{\partial x^k}A^j_mA^k_l\right), $$

or

\begin{align} C^i_{lm} &= \frac{\partial a^i_j}{\partial x^k}A^j_lA^k_m - \frac{\partial a^i_j}{\partial x^k}A^j_mA^k_l\\ &=\partial_m a^i_jA^j_l - \partial_l a^i_jA^j_m. \end{align}